Inverse Matrices

12 Key excerpts on "Inverse Matrices"

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  Precalculus with Limits

  • Ron Larson(Author)
  • 2021(Publication Date)
  • Cengage Learning EMEA

    (Publisher)

  Use a formula to find the inverses of 2 × 2 matrices. Use Inverse Matrices to solve systems of linear equations. The Inverse of a Matrix This section further develops the algebra of matrices. To begin, consider the real number equation ax = b. To solve this equation for x, multiply each side of the equation by a -1 (provided that a ≠ 0). ax = b (a -1 a)x = a -1 b (1)x = a -1 b x = a -1 b The number a -1 is called the multiplicative inverse of a because a -1 a = 1. The multiplicative inverse of a matrix is defined in a similar way. EXAMPLE 1 The Inverse of a Matrix Show that B = [ 1 1 -2 -1 ] is the inverse of A = [ -1 -1 2 1 ] . Solution To show that B is the inverse of A, show that AB = I = BA. AB = [ -1 -1 2 1 ][ 1 1 -2 -1 ] = [ -1 + 2 -1 + 1 2 - 2 2 - 1 ] = [ 1 0 0 1 ] BA = [ 1 1 -2 -1 ][ -1 -1 2 1 ] = [ -1 + 2 -1 + 1 2 - 2 2 - 1 ] = [ 1 0 0 1 ] So, B is the inverse of A because AB = I = BA. This is an example of a square matrix that has an inverse. Note that not all square matrices have inverses. Checkpoint Audio-video solution in English & Spanish at LarsonPrecalculus.com Show that B = [ -1 -3 -1 -2 ] is the inverse of A = [ 2 -3 -1 1 ] . Recall that it is not always true that AB = BA, even when both products are defined. However, if A and B are both square matrices and AB = I n , then it can be shown that BA = I n . So, in Example 1, you need only to check that AB = I 2 . Inverse Matrices are used to model and solve real-life problems. For example, in Exercises 55–58 on page 575, you will use an inverse matrix to find the currents in a circuit. Definition of the Inverse of a Square Matrix Let A be an n × n matrix and let I n be the n × n identity matrix. If there exists a matrix A -1 such that AA -1 = I n = A -1 A then A -1 is the inverse of A. The symbol A -1 is read as “ A inverse.” © Chakrapong Worathat/EyeEm/Getty Images Copyright 2022 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part.

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  Algebra & Trig

  • Ron Larson(Author)
  • 2021(Publication Date)
  • Cengage Learning EMEA

    (Publisher)

  Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 728 Chapter 10 Matrices and Determinants 1.4 Functions GO DIGITAL 10.3 The Inverse of a Square Matrix Verify that two matrices are inverses of each other. Use Gauss-Jordan elimination to find the inverses of matrices. Use a formula to find the inverses of 2 × 2 matrices. Use Inverse Matrices to solve systems of linear equations. The Inverse of a Matrix This section further develops the algebra of matrices. To begin, consider the real number equation ax = b. To solve this equation for x, multiply each side of the equation by a -1 (provided that a ≠ 0). ax = b (a -1 a)x = a -1 b (1)x = a -1 b x = a -1 b The number a -1 is called the multiplicative inverse of a because a -1 a = 1. The multiplicative inverse of a matrix is defined in a similar way. EXAMPLE 1 The Inverse of a Matrix Show that B = [ 1 1 -2 -1 ] is the inverse of A = [ -1 -1 2 1 ] . Solution To show that B is the inverse of A, show that AB = I = BA. AB = [ -1 -1 2 1 ][ 1 1 -2 -1 ] = [ -1 + 2 -1 + 1 2 - 2 2 - 1 ] = [ 1 0 0 1 ] BA = [ 1 1 -2 -1 ][ -1 -1 2 1 ] = [ -1 + 2 -1 + 1 2 - 2 2 - 1 ] = [ 1 0 0 1 ] So, B is the inverse of A because AB = I = BA. This is an example of a square matrix that has an inverse. Note that not all square matrices have inverses. Checkpoint Audio-video solution in English & Spanish at LarsonPrecalculus.com Show that B = [ -1 -3 -1 -2 ] is the inverse of A = [ 2 -3 -1 1 ] . Recall that it is not always true that AB = BA, even when both products are defined. However, if A and B are both square matrices and AB = I n , then it can be shown that BA = I n . So, in Example 1, you need only to check that AB = I 2 . Inverse Matrices are used to model and solve real-life problems. For example, in Exercises 55–58 on page 735, you will use an inverse matrix to find the currents in a circuit.

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  Introduction to Linear Algebra

  A Primer for Social Scientists

  • Gordon Mills(Author)
  • 2017(Publication Date)
  • Routledge

    (Publisher)

  CHAPTER 5The inverse of a square matrix

  5.1 The concept of an inverse matrix

  In Chapter 3 , matrix operations of addition, subtraction and multiplication were defined. In arithmetic and in scalar algebra, there is also an operation of division. Can a similar operation be defined for matrices ? The present chapter deals at some length with this question. The answer is that such an operation can be defined, but the parallel with the division of scalars is by no means exact. In fact the differences are so great that it is better to think in terms of comparing the matrix operation with the scalar operation of multiplying by a reciprocal. In arithmetic, instead of dividing by 2 we can speak of multiplying by (2)−1. . More generally, given a scalar λ≠ 0, we can speak of multiplying by λ −1 ; this operation has the property

  λ

  − 1

  λ = λ

  λ

  − 1

  = 1

  For matrix algebra, this prompts the question: for a given matrix A can a matrix B be found such that

  BA = AB = I

  where I is an identity matrix of order n ? For these matrix equations to hold, B must have n rows and A must have n columns, so that BA is of order n × n. Similarly A must have n rows and B must have n columns if AB is to be of order n×n. In other words if we want to confine our attention to cases where BA and AB are of the same order, then A must be a square matrix, in which case B is also square. This leads to the following definition of an inverse matrix (which bears a rough analogy to the reciprocal of scalar algebra):

  Definition. Given a square matrix A , if there exists a square matrix, to be denoted A−1 , which satisfies the relation

  A

  − 1

  A =

  AA

  − 1

  = I

  then A−1 is called the inverse matrix (or simply the inverse) of A .

  Notice that the definition has to be qualified by the clause ‘if there exists a square matrix’. Nothing that has been said so far demonstrates that such a matrix will exist for a given (square) matrix A . Indeed we shall see later that in some circumstances no such matrix exists, i.e. that it is not possible to find a matrix A −1

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  Elementary Linear Algebra with Supplemental Applications

  • Howard Anton, Chris Rorres(Authors)
  • 2014(Publication Date)
  • Wiley

    (Publisher)

  1.4 Inverses; Algebraic Properties of Matrices 35 As the next theorem shows, identity matrices arise naturally in studying reduced row echelon forms of square matrices. THEOREM 1.4.3 If R is the reduced row echelon form of an n × n matrix A, then either R has a row of zeros or R is the identity matrix I n . Proof Suppose that the reduced row echelon form of A is R = ⎡ ⎢ ⎢ ⎢ ⎣ r 11 r 12 · · · r 1n r 21 r 22 · · · r 2n . . . . . . . . . r n1 r n2 · · · r nn ⎤ ⎥ ⎥ ⎥ ⎦ Either the last row in this matrix consists entirely of zeros or it does not. If not, the matrix contains no zero rows, and consequently each of the n rows has a leading entry of 1. Since these leading 1’s occur progressively farther to the right as we move down the matrix, each of these 1’s must occur on the main diagonal. Since the other entries in the same column as one of these 1’s are zero, R must be I n . Thus, either R has a row of zeros or R = I n . b Inverse of a Matrix In real arithmetic every nonzero number a has a reciprocal a −1 (= 1/a) with the property a · a −1 = a −1 · a = 1 The number a −1 is sometimes called the multiplicative inverse of a. Our next objective is to develop an analog of this result for matrix arithmetic. For this purpose we make the following definition. DEFINITION 1 If A is a square matrix, and if a matrix B of the same size can be found such that AB = BA = I , then A is said to be invertible (or nonsingular ) and B is called an inverse of A. If no such matrix B can be found, then A is said to be singular . Remark The relationship AB = BA = I is not changed by interchanging A and B, so if A is invertible and B is an inverse of A, then it is also true that B is invertible, and A is an inverse of B. Thus, when AB = BA = I we say that A and B are inverses of one another.

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  Technical Mathematics with Calculus

  • Michael A. Calter, Paul A. Calter, Paul Wraight, Sarah White(Authors)
  • 2016(Publication Date)
  • Wiley

    (Publisher)

  In ordinary algebra, if b is some nonzero quantity, then       = = - b b bb 1 1 1 In matrix algebra, the unit matrix I has properties similar to those of the number 1, and the inverse of a matrix has properties similar to the reciprocal in ordinary algebra. ◆◆◆ Example 32: The inverse of the matrix =       = -       - A A 1 0 2 0.5 is 1 0 4 2 1 277 Section 12–3 ◆ The Inverse of a Matrix because =       -       = + - + + - +         =       - AA 1 0 2 0.5 1 0 4 2 1(1) 0( 4) 1(0) 0(2) 2(1) 0.5( 4) 2(0) 0.5(2) 1 0 0 1 1 and = -             = + + - + - +         =       - A A 1 0 4 2 1 0 2 0.5 1(1) 0(2) 1(0) 0(0.5) 4(1) 2(2) 4(0) 2(0.5) 1 0 0 1 1 Not every matrix has an inverse. The inverse exists only for a nonsingular matrix, that is, one whose determinant is not zero. Such a matrix is said to be invertible. Finding the Inverse of a Matrix When we solved sets of linear equations in Chapter 10, we were able to 1. interchange two equations, 2. multiply an equation by a nonzero constant, 3. add a constant multiple of one equation to another equation. Thus, for a matrix that represents such a system of equations, we may perform the following transformations without altering the meaning of a matrix: Elementary Transformations of a Matrix 1. Interchange any rows. 2. Multiply a row by a nonzero constant. 3. Add a constant multiple of one row to another row. 96 One method to find the inverse of a matrix is to apply the rules of matrix transformation to trans- form the given matrix into a unit matrix, while at the same time performing the same operations on a unit matrix. The unit matrix, after the transformations, becomes the inverse of the original matrix. A I I A -1 We will not try to prove the method here, but only show its use. ◆◆◆ Example 33: Find the inverse of the matrix of Example 32.

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  Mathematical Techniques in GIS

  • Peter Dale(Author)
  • 2014(Publication Date)
  • CRC Press

    (Publisher)

  Matrices are made up of numbers such as a ij and a kl that will normally be subject to the four simple operations of addition, subtraction, EXAMPLE 7.5: INVERTING A 4 * 4 MATRIX If A = 3 1 4 2 6 2 1 3 1 2 3 2 4 1 3 7 − then its adjoint A * = 27 27 36 9 60 75 98 43 75 33 1 7 39 12 7 49 − − − − − − − − − AA * = 243 0 0 0 0 243 0 0 0 0 243 0 0 0 0 243 showing the determinant of A = | A | = 243. BOX 7.5 INVERSES AND TRANSPOSES OF SQUARE MATRICES If C = AB , then C T = B T A T and C –1 = B –1 A –1 Similarly, if D = ABC , then D T = C T B T A T and D –1 = C –1 B –1 A –1 where A –1 , and so on, is the inverse of the square matrix A , having the property that AA –1 = A –1 A = I . More generally ( A –1 ) –1 = A ; ( A T ) –1 = ( A –1 ) T ; ( k A ) –1 = k –1 A –1 where k is a sca-lar, A * is the adjugate or adjoint of the square matrix A , made up of the cofac-tors of A in the transposed positions. A * A = | A | I or A * /| A | = A –1 145 Matrices and Determinants multiplication, and division. The answers to their manipulation are then placed in the appropriate location in the new resulting matrix. The processes of manipulating matrices are ideally suited to handling by com-puter as the operations are sequential and routine. Consider a very simple example that involves the intersection of two straight lines illustrated in Box 7.6. The calcula-tion involves the conversion of a square matrix into its inverse ( A into A –1 ) and the multiplication of two matrices ( A –1 * B ). BOX 7.6 THE INTERSECTION OF TWO LINES In Chapter 3, we calculated the intersection of two straight lines. Let us express these in the form a 11 x + a 12 y = b 11 a 21 x + a 22 y = b 21 or = a a a a x y b b 11 12 21 22 11 21 or AX = B Multiply both sides of the expression by A –1 to give A –1 AX = A –1 B .

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  Algebra and Trigonometry

  • Cynthia Y. Young(Author)
  • 2021(Publication Date)
  • Wiley

    (Publisher)

  STUDY TIP • Only a square matrix can have an inverse. • Not all square matrices have inverses. The identity matrix I n will assist us in developing the concept of an inverse of a square matrix. It is important to note that only a square matrix can have an inverse. Even then, not all square matrices have inverses. EXAMPLE 2 Multiplying a Matrix by the Multiplicative Identity Matrix I n For A = [ −2 4 1 3 7 −1 ] , find I 2 A. Solution Write the two matrices. A = [ −2 4 1 3 7 −1 ] I 2 = [ 1 0 0 1 ] Find the product I 2 A. I 2 A = [ 1 0 0 1 ][ −2 4 1 3 7 −1 ] I 2 A = [ (1)(−2) + (0)(3) (1)(4) + (0)(7) (1)(1) + (0)(−1) (0)(−2) + (1)(3) (0)(4) + (1)(7) (0)(1) + (1)(−1) ] I 2 A = [ −2 4 1 3 7 −1 ] = A Your Turn For A in Example 2, find AI 3 . Answer A I 3 = [ −2 4 1 3 7 −1 ] = A 982 CHAPTER 10 Matrices EXAMPLE 3 Multiplying a Matrix by Its Inverse Verify that the inverse of A = [ 1 3 2 5 ] is A −1 = [ −5 3 2 −1 ] . Solution Show that A A −1 = I 2 and A −1 A = I 2 . Find the product A A −1 . A A −1 = [ 1 3 2 5 ][ −5 3 2 −1 ] = [ (1)(−5) + (3)(2) (1)(3) + (3)(−1) (2)(−5) + (5)(2) (2)(3) + (5)(−1) ] = [ 1 0 0 1 ] = I 2 Find the product A −1 A. A −1 A = [ −5 3 2 −1 ][ 1 3 2 5 ] = [ (−5)(1) + (3)(2) (−5)(3) + (3)(5) (2)(1) + (−1)(2) (2)(3) + (−1)(5) ] = [ 1 0 0 1 ] = I 2 Your Turn Verify that the inverse of A = [ 1 4 2 9 ] is A −1 = [ 9 −4 −2 1 ] . Answer AA −1 = A −1 A = I 2 Now that we can show that two matrices are inverses of one another, let us describe the process for finding an inverse, if it exists. If an inverse A −1 exists, then the matrix A is said to be nonsingular. If the inverse does not exist, then the matrix A is said to be singular. Let A = [ 1 −1 2 −3 ] and the inverse be A −1 = [ w x y z] , where w, x, y, and z are variables to be determined. A matrix and its inverse must satisfy the identity A A −1 = I 2 . Words Math The product of a matrix and [ 1 −1 2 −3 ][ w x y z] = [ 1 0 0 1 ] its inverse is the identity matrix.

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  Fundamental Maths

  For Engineering and Science

  • Mark Breach(Author)
  • 2017(Publication Date)
  • Red Globe Press

    (Publisher)

  30-09 Matrix inverse The product of any square matrix and its inverse, in either order, is an identity matrix of the same dimensions: A 2 A 1 2 ¼ I 2 A 3 A 1 3 ¼ I 3 A n A 1 n ¼ I n The inverse of a two-by-two matrix is found by: transposing elements a 1 ; 1 and a 2 ; 2 ; multiplying elements a 1 ; 2 and a 2 ; 1 by 1; dividing the new matrix by the determinant of the original matrix. For example, if A ¼ a 1 ; 1 a 1 ; 2 a 2 ; 1 a 2 ; 2 then A 1 ¼ 1 j A j a 2 ; 2 a 1 ; 2 a 2 ; 1 a 1 ; 1 Now, form the product to verify that AA 1 ¼ I : A 1 A ¼ 1 j A j a 2 ; 2 a 1 ; 2 a 2 ; 1 a 1 ; 1 a 1 ; 1 a 1 ; 2 a 2 ; 1 a 2 ; 2 ¼ 1 j A j a 2 ; 2 a 1 ; 1 a 1 ; 2 a 2 ; 1 a 2 ; 2 a 1 ; 2 a 1 ; 2 a 2 ; 2 a 2 ; 1 a 1 ; 1 þ a 1 ; 1 a 2 ; 1 ¼ a 2 ; 1 a 1 ; 2 þ a 1 ; 1 a 2 ; 2 ¼ 1 j A j j A j 0 0 j A j ¼ 1 0 0 1 ¼ I Finding the inverse of a two-by-two matrix is a scaled-down version of the procedure for finding the inverse of a larger matrix. This is a little more involved and requires that we introduce some new terms. A minor of a matrix is the determinant of the original matrix once the row and column containing a subject element have been removed. For example, in 5 7 9 8 3 2 6 1 4 2 4 3 5 if the subject element is the 8 – second row and first column – the determinant of this matrix with the second row and first column removed is 7 4 9 1 ¼ 19. Hence this minor of the matrix is 19. A signed minor is a minor multiplied by ð 1 Þ i þ j where i is the row number and j is the column number of the subject element. So if i þ j is even there is no change to the minor. If i þ j is odd the minor changes sign. In the above example, the 8 is in the second row and first column so the signed minor is multiplied by ð 1 Þ 2 þ 1 ¼ð 1 Þ 3 ¼ 1 to give 19. We also need the adjoint of a matrix in order to find the inverse of a matrix. The dimensions of the adjoint are the same as those of the original matrix.

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  Linear Algebra, Geometry and Transformation

  • Bruce Solomon(Author)
  • 2014(Publication Date)
  • Chapman and Hall/CRC

    (Publisher)

  If it is, find its inverse. a)     1 1 1 1 1 1 -1 -1 1 -1 -1 1 1 -1 1 -1     b)     2 1 1 1 1 0 -1 -1 1 -1 0 1 1 -1 1 2     235 . Show that the general 2 × 2 matrix A = a b c d is invertible if and only if ad -bc 6 = 0, in which case A -1 is given by A -1 = 1 ad -bc d -b -c a This general formula for the inverse of a 2 × 2 matrix is worth memo-rizing. 4. A LOGICAL DIGRESSION 199 236 . For what values of λ (if any) are the matrices below singular, i.e., non -invertible? a) λ 3 3 λ b) 1 -λ 5 5 1 -λ c) 1 -λ 5 0 2 -λ d) λ 1 -1 λ 237 . Prove these easy properties of the matrix inverse. a) ( A -1 ) -1 = A b) If A is invertible, then so is k A for any scalar k 6 = 0, with ( k A ) -1 = 1 k A -1 c) If A is invertible, so is A p for any positive integer p , with ( A p ) -1 = ( A -1 ) p 238 . Proposition 3.10 tells us that the product of invertible n × n matrices is again invertible. Is the same true for sums? That is, does invertibility of A and B imply that of A + B ? Prove it does, or show (by counterexample) that it does not. 239 . If an upper-triangular matrix is invertible, its inverse will also be upper-triangular. Can you explain this as a consequence of the inversion algorithm? — ? — 4. A Logical Digression We now want to understand the inversion algorithm—to see why it works. For this, and increasingly as we go forward, we will use mathe-matical reasoning at least as much as calculation . To prepare for that, we digress to briefly discuss some basic facts about logic. Specifically, we present the four fundamental templates for the relation-ship between a pair of mathematical claims, or assertions P and Q . These four templates are known as statement , converse , obverse , and contrapositive . 200 4. THE ALGEBRA OF MATRICES 4.1. Statements. Every mathematical theorem makes a state-ment about some set of objects, where by statement , we specifically mean an assertion of the form If P then Q This is the same as saying P implies Q .

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  Invitation to Linear Programming and Game Theory

  • David C. Vella(Author)
  • 2021(Publication Date)
  • Cambridge University Press

    (Publisher)

  Then the inverse of A is given by the formula A −1 = 1 det(A) A adj (2.11) = 1 ad − bc  d −b −c a  . That is, simply multiply the adjoint matrix by a scalar factor equal to the reciprocal of the determinant. For example, with A =  2 3 5 −1  , we have det(A) = −17, so A −1 = 1 −17  −1 −3 −5 2  =  1 17 3 17 5 17 − 2 17  . This checks using the definition of inverse: AA −1 =  2 3 5 −1   1 17 3 17 5 17 − 2 17  =  1 0 0 1  = I 2 as expected, and similarly for the product A −1 A, as the reader should verify. Again, with proper definition of adjoints and determinants for larger matrices, this theorem extends to n × n matrices. We eschew the larger case here, but the n = 2 case is easy to prove, and we do so now: Proof We must check that AA −1 = I 2 = A −1 A. But observe, assuming det A = ad − bc = 0, we have by Theorem 2.6, Part d: 2.3 Operations on Matrices II: Matrix Inversion 81 AA −1 = A  1 det A A adj  = 1 det A AA adj = 1 det A  a b c d   d −b −c a  = 1 det A  ad − bc 0 0 ad − bc  = 1 ad − bc  ad − bc 0 0 ad − bc  =  1 0 0 1  = I 2 . The product in the other order is similar and left for the reader to check. This concludes the proof.  As a corollary, we have a proof of part of Theorem 2.14. Indeed, if det A = 0, then Equation (2.11) exhibits an inverse of A. The reverse implication (the “only if” part of the statement) would say that if det A = 0, then no inverse exists. Clearly, Formula 2.11 on the facing page cannot work if det A = 0, since we would be dividing by zero. However, that doesn’t prove that some other matrix could not be the inverse in this case. In the exercises, the reader will prove the “only if” implication by using a known property of the determinant. The next theorem summarizes some properties of the inverse. theorem 2.16 Let A be an invertible n × n matrix. Then: a. The inverse of A is unique. b. ( A −1 ) −1 = A c. A T is also invertible, and ( A T ) −1 = ( A −1 ) T d.

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  Advanced Engineering Mathematics

  • Lawrence Turyn(Author)
  • 2013(Publication Date)
  • CRC Press

    (Publisher)

  Theorem 1.30 is one of the mathematical results most often used by engineers and scientists. Because of Theorem 1.30, we know the matrix A is invertible in each of Examples 1.36 and 1.37. Theorem 1.30 is an “existential result,” that is, it tells us whether A − 1 exists but does not tell us how to find A − 1 . But Theorem 1.30 illustrates the concept that if we study the explanation of a theorem, we can discern an algorithm for getting numerical results. Here, the row reduction algorithm for determinants is what makes the explanation “tick.” Theorem 1.31 If A − 1 exists, then | A − 1 | = 1 | A | . Besides using elementary row operations to evaluate a determinant, we can also use “ elementary column operations ” because of the identity | A | = | A T | . For example, the elementary column operation of interchanging two columns given by C i ↔ C j , multiplies a determinant by ( − 1 ) because the operation of interchanging two rows of A T multiplies its determinant by ( − 1 ) . Similarly, the operation of adding a multiple of one column into another column does not change the determinant, for example, 2 C 1 + C 3 → C 3 doesn’t affect the determinant. Linear Algebraic Equations, Matrices, and Eigenvalues 53 1.6.1 Adjugate Matrix Definition 1.27 The adjugate of A is defined by adj ( A ) A ji 1 ≤ j ≤ n 1 ≤ i ≤ n = A ij 1 ≤ i ≤ n 1 ≤ j ≤ n T , that is, adj ( A ) is the transpose of the matrix of cofactors of A . By the way, many people call adj ( A ) the “adjoint of A ” but use the same abbreviation, “adj.” We prefer to not use the word adjoint because it means something entirely differ-ent in the subject of operator theory, which has many applications to matrix theory and differential equations. Theorem 1.32 (a) A adj ( A ) = | A | I n = adj ( A ) A . (b) If | A | = 0, then A − 1 = 1 | A | adj ( A ) . (c) If | A | = 0, then A adj ( A ) = O = adj ( A ) A . Theorem 1.32(b) gives a formula for A − 1 that can be useful for very small matrices and also for theoretical purposes.

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  College Algebra

  • Cynthia Y. Young(Author)
  • 2021(Publication Date)
  • Wiley

    (Publisher)

  A −1 A = I n STEP 1 Form the matrix [A | I n ]. STEP 2 Use row operations to transform this matrix to [I n | A −1 ]. Note: Not every square matrix has an inverse. Solving systems of linear equations using matrix algebra and inverses of square matrices AX = B STEP 1 Find A −1 . STEP 2 X = A −1 B. 7.4 The determinant of a square matrix and Cramer’s rule Cramer’s rule can be used to solve only a square system of linear equations with a unique solution. Determinant of a 2 × 2 matrix | a b c d| = ad − bc Determinant of an n × n matrix Let A be a square matrix of order n × n; then • The minor M ij of the element a ij is the determinant of the (n − 1) × (n − 1) matrix obtained when the ith row and jth column of A are deleted. • The cofactor C ij of the element a ij is given by C ij = (−1) i+j M ij . ⎡ ⎢ ⎣ 1 −3 2 4 −1 0 5 −2 3 ⎤ ⎥ ⎦ M 11 = | −1 0 −2 3| = −3 − 0 = −3 C 11 = (−1) 1+1 M 11 = (1)(−3) = −3 Sign Array of a 3 × 3 matrix: [ + − + − + − + − + ] If A is a 3 × 3 matrix, the determinant can be given by det(A) = a 11 C 11 + a 12 C 12 + a 13 C 13 . This is called expanding the determinant by the first row. (Note that any row or column can be used.) | a 1 b 1 c 1 a 2 b 2 c 2 a 3 b 3 c 3| = a 1 | b 2 c 2 b 3 c 3| − b 1 | a 2 c 2 a 3 c 3| + c 1 | a 2 b 2 a 3 b 3| Cramer’s rule: Systems of linear equations in two variables The system a 1 x + b 1 y = c 1 a 2 x + b 2 y = c 2 has the solution x = D x ___ D y = D y ___ D if D ≠ 0 where D = | a 1 b 1 a 2 b 2 | D x = | c 1 b 1 c 2 b 2 | D y = | a 1 c 1 a 2 c 2| Chapter 7 Review 627 Section Concept Key Ideas/Formulas Cramer’s rule: Systems of linear equations in three variables The system a 1 x + b 1 y + c 1 z = d 1 a 2 x + b 2 y + c 2 z = d 2 a 3 x + b 3 y + c 3 z = d 3 has the solution x = D x ___ D y = D y ___ D z = D z ___ D if D ≠ 0 where D = | a 1 b 1 c 1 a 2 b 2 c 2 a 3 b 3 c 3 | D x = | d 1 b 1 c 1 d 2 b 2 c 2 d 3 b 3 c 3 | D y = | a 1 d 1 c 1 a 2 d 2 c 2 a 3 d 3 c 3 | D z = | a 1 b 1 d 1 a 2 b 2 d 2 a 3 b 3 d 3 |

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